Question

A wire of length 0.050 m and carries a current I is in a horizontal uniform magnetic field B of 2.50 T which is perpendicular to the rod. The weight of the wire is 0.80 N. What is the value of the current I if the magnetic force on the wire equals its weight?

Answer 1

According to Ampere's law, absolute value of force, acting on the rod is $F = B I l \sin \alpha$, where $\alpha$ is the angle between the direction of current, and magnetic field. Since $\alpha = 90^{\circ}$, $F = B I l = P$, from where $I = \frac{P}{B l} = \frac{0.80 N}{2.5 T \cdot 0.05 m} \approx 6.4 A$